CHEMISTRY COMMUNITY

Naomi 3G

I would really suggest watching the video modules and then doing the post module assessment. The videos explain the concepts really clearly and you can pause them to take notes or think through concepts. They also have clear examples of worked problems and then the assessments are really great to se...

Naomi 3G

Can someone explain this question in the video model assessment. Molybdenum metal must absorb radiation with a minimum frequency of 1.09*10^15s^-1 before is can emit an electron from its surface. The minimum energy needed to produce this effect id 7.22*10^-19J If molybdenum is irradiated with 194 nm...

Naomi 3G

Could someone explain to be how you use the work function to find kinetic energy of an electron and the energy needed to remove an electron. Specifically in the case of these two questions from the video module assessments. "Light hits a sodium metal surface and the velocity of the ejected elec...

Naomi 3G

I wanted to say thank you to Dr. Lavelle for creating such helpful video modules for each topic in this course. Lots of times in class it's hard for me to comprehend everything that is being taught because I don't have enough time to process each point. However, the videos allow me to go through the...

Naomi 3G

Can someone clarify why the Rydberg equation is always negative? Because you're "losing" energy because it's being transferred somewhere else? The Rydberg equation is negative because it is looking at the change in energy when you go from a high state to a lower state. If you are going fr...

Naomi 3G

Actually, I think the book uses c to represent molarity and M to represent molar mass in equations. I was a bit confused by this too, since M is the symbol for units in molarity (ie .5 M). So in the equation m=cMV, it's basically mass=molarity x molar mass x volume. Ya, I agree, in the book M in eq...

Naomi 3G

Is there somewhere where we can see what this sheet will look like before we take the quiz so that we can get used to what equations we will have and where they will be.

Naomi 3G

Hi,

I am also confused why in the solutions book for a they multiply .07977 by 2 in the denominator of the equation

Naomi 3G

Chem_Mod wrote:Here is the answers for today's workshop.

Thanks so much, this was super helpful!

CHEMISTRY COMMUNITY

Search found 9 matches

Re: Best Way To Study?

Video Module Assessment

Re: The Work Function

Re: Saying Thank You to Dr. Lavelle

Re: Rydberg equation [ENDORSED]

Re: Symbols for Molarity

Re: Periodic table

Re: G5

Re: KAREN SUN 5-7PM WORKSHOP - DOWNLAOD WORKSHEETS HERE